# 1963 AHSME Problems/Problem 33

## Problem 33

Given the line $y = \dfrac{3}{4}x + 6$ and a line $L$ parallel to the given line and $4$ units from it. A possible equation for $L$ is: $\textbf{(A)}\ y =\frac{3}{4}x+1\qquad \textbf{(B)}\ y =\frac{3}{4}x\qquad \textbf{(C)}\ y =\frac{3}{4}x-\frac{2}{3}\qquad \\ \textbf{(D)}\ y = \dfrac{3}{4}x -1 \qquad \textbf{(E)}\ y = \dfrac{3}{4}x + 2$

## Solution $[asy] import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((16/3,10)--(-10,-3/2),Arrows); dot((12/5,14/5)); dot((0,6)); draw((0,6)--(12/5,14/5),dotted); [/asy]$

If a point on line $L$ is four units away from another line, the perpendicular distance between the two lines is $4$. Since a line perpendicular to the original has a slope of $-\tfrac{4}{3}$, a point on that line is $5d$ units away and can be derived from subtracting $4d$ from the y-coordinate and adding $3d$ to the x-coordinate of a point on the original line.

We want the perpendicular distance to be $4$, so $d = \tfrac{4}{5}$. Since one point on the original line is $(0,6)$, a point on line $L$ will be $(0+3 \cdot \tfrac{4}{5},6 - 4 \cdot \tfrac{4}{5})$, which is simplified as $(\tfrac{12}{5},\tfrac{14}{5})$. Using point-slope form, the equation of line $L$ is $$y - \frac{14}{5} = \frac{3}{4} (x - \frac{12}{5})$$ $$y = \frac{3}{4} x + 1$$ The answer is $\boxed{\textbf{(A)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 